A satellite used in a cellular telephone network has a mass of 2380 kg and is in a circular orbit at a height of 850 km above the surface of the earth. |
Part A What is the gravitational force Fgrav on the satellite? Take the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , the mass of the earth to be me = 5.97×1024 kg , and the radius of the Earth to be re = 6.38×106 m |
A satellite used in a cellular telephone network has a mass of 2380 kg and is...
A satellite used in a cellular telephone network has a mass of 2050 kg and is in a circular orbit at a height of 880 km above the surface of the earth. Part A What is the gravitational force Fgrav on the satellite? Take the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , the mass of the earth to be me = 5.97×1024 kg , and the radius of the Earth to be re = 6.38×106 m . part...
A 2830 kg satellite used in a cellular telephone network is in a circular orbit at a height of 740 km above the surface of the earth. What is the gravitational force on the satellite? What fraction is this force of the satellite's weight at the surface of the earth?
A 2660 kg satellite used in a cellular telephone network is in a circular orbit at a height of 720 km above the surface of the earth. What is the gravitational force on the satellite?(Answer in N). What fraction is this force of the satellite's weight at the surface on the Earth?(Answer in %).
A 2710 kg satellite used in a cellular telephone network is in a circular orbit at a height of 790 km above the surface of the earth. part a: What is the gravitational force on the satellite? part b: What fraction is this force of the satellite's weight at the surface of the earth? (convert to %)
A satellite m = 500 kg orbits the earth at a distance d = 218 km, above the surface of the planet. The radius of the earth is re = 6.38 × 106 m and the gravitational constant G = 6.67 × 10-11 N m2/kg2 and the Earth's mass is me = 5.98 × 1024 kg. What is the speed of the satellite in m/s?
A 2560 kg satellite used in a cellular telephone network is in a circular orbit at a height of 720 kmabove the surface of the earth. What is the gravitational force on the satellite? What fraction is this force of the satellite's weight at the surface of the earth?
A 2250 kg satellite used in a cellular telephone network is in a circular orbit at a height of 740 kmabove the surface of the earth. A) What is the gravitational force on the satellite? B) What fraction is this force of the satellite's weight at the surface of the earth?
A 2160 kg satellite used in a cellular telephone network is in a circular orbit at a height of 800 km above the surface of the earth. What is the gravitational force on the satellite? What fraction is this force of the satellite's weight at the surface of the earth? I've looked at other problems but I'm not getting the correct answer, I've gotten 21168 for the first one and 0.998 for the second part.
A 2940 kg satellite used in a cellular telephone network is in a circular orbit at a height of 770 kmabove the surface of the earth. Part A What is the gravitational force on the satellite? Part B What fraction is this force of the satellite's weight at the surface of the earth?
Consider a 475 kg satellite in a circular orbit at a distance of 3.06 x 104 km above the Earth's surface. What is the minimum amount of work W the satellite's thrusters must do to raise the satellite to a geosynchronous orbit? Geosynchronous orbits occur at approximately 3.60 x 104 km above the Earth's surface. The radius of the Earth and the mass of the Earth are RE = 6.37 x 109 km and Me = 5.97 x 1024 kg,...